Such attempted generalizations of ABC to four or more variables often fail to specializations of the identity $$ (x^2+xy-y^2)^3 + (x^2-xy-y^2)^3 = 2 (x^6 - y^6). \label{1}\tag{*} $$ One can use elliptic curves to make both $x^2 + xy - y^2$ and $x^2 - xy - y^2$ "powerful" (of the form $A^2 B^3$), which makes each of the four terms $(x^2+xy-y^2)^3$, $(x^2-xy-y^2)^3$, $2x^6$, $2y^6$ have $h=6$ but for a stray factor of $2$ which should not matter in the context of the ABC conjecture. For example, the pairwise prime numbers $a,b,c,d$ below satisfy $2a^6 + b^6 + 61^9 c^6 = 2d^6$. Here $d$ is even but $a$ is odd, so $2a^6$ has a "stray factor of $2$", and the expansion to $a^6 + a^6 + b^6 + 61^9 c^6 = 2d^6$ loses pairwise coprimality; so either way we don't quite get a counterexample. Still, this suggests that generalizations of ABC to four or more variables can run afoul of identities such as \eqref{1}. (It is "well known" that the Mason-Stothers theorem forbids the disproof of ABC itself by such an identity.)
a = 1022288301691921314835532892967014277786302791344455107816139963763145069687359424810667270039489345929029301393007247303344511065237
b = 4005821025365458069945118311017282675402206671149976403498624129498574702167905733126870212117037684063261425637225699359421949547271
c = 10621830276852061412855232703108032231130723932854745057900981539571749281974534306702514113168069346943754838515856358759614674721
d = 4676830625123658957500070687744472849236478810555581279857582626862200039312130562436302012081022720213179152015505627679021327325170
